Past exam of the options I course of the University of Oxford / 2009 / M / M2 / b / Solution

Let $u_n=\mathbb{P}(X_n=1\mid X_0=1)$. Conditioning on $X_n$ givesThe fixed value is $b/(a+b)$. Subtracting it and iterating the scalar recurrence yieldsThe formula includes $n=0$. If $a+b=1$, its value at $n=0$ is interpreted directly as $u_0=1$, while $u_n=b$ for all $n\ge 1$. Since $|1-a-b|<1$, the limiting probability is $b/(a+b)$.